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clang 3.6 fold expression left/right

I'm trying the fold expression with clang 3.6 '--std=c++1z', but something I don't quite get. The function that I'm testing is:

auto minus = [](auto... args) { return (args - ...); };
...
std::cout << minus(10, 3, 2) << std::endl;

according to n4191, I'm expecting it expands as a left fold to

(10 - 3) - 2

which gives result 5, however, the result is 9, which seems to be a right fold expansion, i.e.

10 - (3 - 2)

Am I missing anything or mis-understand n4191? Thanks

like image 895
Ralph Zhang Avatar asked Mar 02 '15 08:03

Ralph Zhang


1 Answers

n4191 was revised by n4295. According to that, an expression of the form (e op ...) is a unary right fold, and that is expanded as: E1 op (... op (EN-1 op EN)), i.e. as a right fold expansion.

This does seem to be the reverse of what n4191 stated in terms of the fold direction. Clang 3.6 implements the n4295 proposal, as shown here.

... - args would be a unary left fold and expand in the direction you want.

like image 139
interjay Avatar answered Nov 02 '22 16:11

interjay